Fundamentholfundamenthol

Preparation of Alkanes

ChemistryHydrocarbonsFor JEE aspirants

The preparation of alkanes falls into three groups decided by what happens to the carbon count. Reduction methods keep it the same, coupling methods such as the Wurtz reaction double it, and decarboxylation drops it by one. This page covers every route in the JEE and NEET syllabus for the preparation of alkanes, with the mechanism, the conditions and the traps for each one. Learn the carbon bookkeeping first and the reagents become far easier to recall.

Same carbon count7 routesreductions and hydrolysis
More carbons4 routesWurtz, Frankland, Corey-House, Kolbe
One carbon fewer1 routesoda lime decarboxylation
First move in any questionCount the carbonsit narrows the method instantly
Key Formulas - Quick Reference
  1. Hydrogenation:
  2. Halide reduction:
  3. Wurtz:
  4. Frankland:
  5. Corey-House:
  6. Kolbe:
  7. Decarboxylation:
  8. Grignard:

1The Map: Sort Every Method by Carbon Count

There are about a dozen named ways to make an alkane, and students lose marks by mixing them up. The organising question is simple: does the product have the same number of carbons as the starting material, twice as many, or one fewer?

Map of all preparation routes to alkanes grouped by carbon count Three columns of preparation methods converging on an alkane. The first column keeps the carbon count unchanged and holds hydrogenation of alkenes and alkynes and reduction of alkyl halides, alcohols, carbonyl compounds and acids. The second column doubles the carbon count and holds the Wurtz reaction, Frankland reaction, Corey House synthesis, Kolbe electrolysis and hydroboration coupling. The third column reduces the count by one and holds decarboxylation with soda lime. Every route to an alkane, sorted by what it does to the carbon count n carbons in, n out Hydrogenation of alkenes and alkynes Reduction of alkyl halides Reduction of alcohols Reduction of aldehydes and ketones Reduction of acids by HI/P n carbons in, 2n out Wurtz reaction Frankland reaction Corey-House synthesis (gives unsymmetrical R-R') Kolbe electrolysis Hydroboration coupling Grignard with R'X n in, n-1 out Decarboxylation of the sodium salt of an acid with soda lime Special case Hydrolysis of carbides gives methane only Berthelot: C + H₂ in an arc Alkane R—H
Fig 1The single most useful way to hold these methods in memory. Ask what happens to the carbon count: unchanged, doubled to , or dropped to .

2Reduction Methods (Carbon Count Unchanged)

2.1 Hydrogenation of alkenes and alkynes

Alkenes and alkynes add hydrogen across the multiple bond in the presence of a finely divided catalyst such as nickel, platinum or palladium. This is the Sabatier and Senderens reaction.

Preparation of alkanes by catalytic hydrogenation of alkenes and alkynes Two reaction rows. In the first, an alkene with a carbon carbon double bond takes up one mole of hydrogen over nickel at 250 degrees to give an alkane. In the second, an alkyne with a triple bond takes up two moles of hydrogen over platinum to give the same alkane. The carbon skeleton is unchanged in both. R R' alkene + H2 Ni, 250°C Pt or Pd also work R R' alkane R R' alkyne + 2H2 Pt two moles of H₂ needed R R' same alkane Catalytic hydrogenation is also called the Sabatier and Senderens reaction
Fig 2Catalytic hydrogenation. An alkene needs one , an alkyne needs two, and neither changes the carbon skeleton.
An alkyne needs two moles of to reach the alkane, an alkene only one. Nickel needs heating to about ; platinum and palladium work at room temperature.
How a nickel, palladium or platinum catalyst adds hydrogen to a double bond Left panel shows the metal catalyst surface. Hydrogen molecules stick to the metal and the hydrogen hydrogen bond breaks into two separate hydrogen atoms held on the surface. The alkene also sticks to the surface through its pi bond. Right panel shows the result. Both hydrogen atoms are delivered from the same face of the double bond, so catalytic hydrogenation is syn or cis addition, and the alkane then leaves the surface. Why the metal is needed: the surface splits H2 1. Both partners stick to the metal 2. Both H arrive on the same face Ni / Pd / Pt SURFACE H H H H H-H breaks C C π bond alkene sticks here Ni / Pd / Pt SURFACE C C H H new single bond alkane leaves syn (cis) addition – same face Activity order: Pt > Pd > Ni. Ni is cheapest, so it is used industrially but needs 250-300°C. This route never changes the carbon count.
Fig 3Catalytic hydrogenation happens on the metal, not in solution. The surface breaks and holds both atoms, so they add to the same face of the bond. That is why the addition is syn, and why finely divided metal (large surface area) works best.

2.2 Reduction of alkyl halides

Alkyl halides undergo reduction with nascent hydrogen to give alkanes. The halogen is simply swapped for hydrogen, so the skeleton survives intact.

Reagents that reduce an alkyl halide to an alkane A fan diagram with R X on the left and R H on the right, joined by eight reagent lines. The reagents are lithium aluminium hydride, sodium borohydride, triphenyltin hydride, palladium on carbon with hydrogen, sodium in ethanol, zinc copper couple in ethanol, zinc with sodium hydroxide or acetic acid, and zinc amalgam with water. The carbon count is unchanged. R—X n carbons R—H n carbons LiAlH4 NaBH4 Ph3SnH (TPH) Pd-C / H2 Na / C2H5OH Zn-Cu couple / C2H5OH Zn / NaOH or Zn / CH3COOH Zn-Hg / HOH R—X + 2[H] → R—H + HX, a simple replacement of X by H
Fig 4Eight ways to turn into . All replace the halogen with hydrogen, so the carbon count never changes.
Why lithium aluminium hydride fails with tertiary alkyl halides Tertiary butyl chloride treated with lithium aluminium hydride gives 2-methylpropene rather than isobutane. Lithium aluminium hydride is a strong base as well as a hydride donor, so with a tertiary halide it causes elimination instead of reduction. H3C CH3 CH3 Cl tert-butyl chloride, a 3° halide LiAlH4 H2C CH3 CH3 2-methylpropene, not an alkane LiAlH₄ is a strong base as well as a hydride donor with a 3° alkyl halide it eliminates instead of reducing, so you get an alkene
Fig 5The exception worth memorising. reduces and halides cleanly, but eliminates with a halide to give an alkene.
  • reduces and halides well, but it is also a strong base, so with a halide it gives an alkene by elimination
  • with water and with acetic acid are the mild options when other groups must survive
  • works but will also reduce any double bond present in the molecule
SOURCE CHECK Some coaching notes state that reduces only and alkyl halides. Treat that with care. delivers hydride by an pathway, which favours and substrates; halides are reduced only under special ionising conditions. For exam purposes, the safe and repeatedly tested statement is the one above.

2.3 Reduction of alcohols

Two routes from an alcohol to an alkane An alcohol R O H can be converted to the alkane R H directly by red phosphorus and iodine on heating, or in two steps by first making the tosylate R O Ts with tosyl chloride and then reducing with lithium aluminium hydride. The hydroxyl group is a poor leaving group, which is why the tosylate route exists. R—OH alcohol, n carbons P / I2 / Δ one step, direct R—H alkane, n carbons TsCl tosyl chloride R—O—Ts LiAlH4 R—H —OH is a poor leaving group, so it is first converted into the tosylate —OTs —OTs leaves easily, and hydride from LiAlH₄ then takes its place
Fig 6Alcohols reach the alkane either directly with , or through the tosylate, because itself is a poor leaving group.

2.4 Reduction of aldehydes and ketones

The carbonyl oxygen is removed entirely and replaced by two hydrogens, so becomes . Three reagent systems do this, and the choice between them is a favourite exam point.

Clemmensen and Wolff-Kishner reduction of carbonyl compounds to alkanes An aldehyde or ketone is reduced to an alkane by three routes. Zinc amalgam with concentrated hydrochloric acid is the Clemmensen reduction and works in acid. Hydrazine with dilute potassium hydroxide on heating is the Wolff Kishner reduction and works in base. Hydroiodic acid with red phosphorus on heating also works. The carbonyl oxygen is replaced by two hydrogens. O C R R' aldehyde or ketone Zn-Hg / conc. HCl Clemmensen reduction, acidic NH2—NH2 / dil. KOH / Δ Wolff-Kishner reduction, basic HI / red P / Δ strongly reducing, acidic R—CH2—R' alkane Choose by what else is in the molecule: Clemmensen for acid-stable substrates, Wolff-Kishner when the molecule cannot survive strong acid
Fig 7The carbonyl group becomes . Clemmensen runs in acid, Wolff-Kishner in base, so they cover each other's weaknesses.
ReductionReagentMediumUse when
Clemmensen / conc. Strongly acidicThe molecule tolerates acid
Wolff-Kishner / dil. , BasicThe molecule is acid-sensitive
Red + / red / Strongly acidicA powerful, non-selective reduction is acceptable

2.5 Reduction of carboxylic acids

Heating a carboxylic acid with and red phosphorus reduces all the way to without losing a carbon.

Contrast this with decarboxylation in Section 4. Both start from an acid, but keeps all carbons while soda lime removes one. Questions are built on exactly this distinction.

3From Organometallic Compounds

Grignard reagents and organolithiums carry a strongly nucleophilic, basic carbon. Give that carbon any acidic hydrogen, from water, alcohol, ammonia or an acid, and it is protonated straight to the alkane.

Alkanes from Grignard reagents and organolithiums A Grignard reagent has a partial negative charge on carbon and a partial positive charge on magnesium, so the carbon takes a proton from water to give an alkane and a basic magnesium halide. An organolithium behaves the same way. Using heavy water in place of water gives a deuterated alkane. Using an alkyl halide instead of water couples the two groups to give a larger alkane. R—MgX δ− δ+ + H—OH δ+ δ− R—H + Mg(OH)X R—Li + H—OH R—H + LiOH Exam favourite: swap the water for heavy water (CH3)2CH—MgBr + D2O (CH3)2CH—D Give the Grignard an alkyl halide instead of water and it couples: R—MgX + R'—X → R—R', a larger alkane
Fig 8The carbon of a Grignard reagent carries charge, so any acidic hydrogen converts it straight to the alkane. Use and you get .
Formation of a Grignard reagent and its hydrolysis to an alkane Step one, an alkyl halide reacts with magnesium in dry ether to give the Grignard reagent R Mg X. The carbon carries a partial negative charge and behaves like a carbanion, while magnesium carries a partial positive charge. Step two, that carbanion carbon grabs a proton from water, giving the alkane R H and a basic magnesium salt. A panel lists the other proton sources that do the same job, water, alcohols, ammonia, carboxylic acids and mineral acids, which is why the apparatus must be completely dry. The Grignard route: make a carbanion, then feed it a proton STEP 1 Dry ether is essential – a trace of water kills the reagent R–X + Mg dry ether reflux R–Mg–X δ− δ+ carbon is the nucleophile STEP 2 The carbanion takes a proton and becomes the alkane R–Mg–X + H–OH C grabs the H R–H + Mg(OH)X the alkane ANY OF THESE WILL DESTROY A GRIGNARD REAGENT H–OH R–OH NH3 RCOOH H–X water alcohol ammonia carboxylic acid mineral acid Each one has a hydrogen bonded to O, N or halogen, so each one is acidic enough.
Fig 9The Grignard reagent works because the polarity of the bond leaves carbon with a charge. That carbanion-like carbon pulls a proton off anything even slightly acidic, giving . The same property is why the glassware must be bone dry.

4Coupling Reactions (Carbon Count Doubled)

4.1 Wurtz reaction

The Wurtz reaction and its four limitations When two identical alkyl halides react with sodium in dry ether, a single alkane with twice the carbon count is formed. When two different alkyl halides are used, three products form and are hard to separate. The reaction cannot make methane, fails with tertiary halides, produces alkenes as by-products, and only gives alkanes with an even number of carbons. Case 1: both alkyl halides the same, one clean product R—X + R—X Na, dry ether R—R + 2NaX 2n carbons Case 2: different alkyl halides, three products and a separation problem R—X + R'—X Na, dry ether R—R + R'—R' + R—R' Example: C₂H₅Br with C₄H₉Br gives butane, hexane and octane together Four limitations that are asked directly Methane cannot be made this way, since the smallest product is ethane 3° alkyl halides fail, they eliminate instead of coupling Alkenes appear as by-products, and only even-carbon alkanes are accessible
Fig 10The Wurtz reaction doubles the carbon count. It is clean only when both halides are identical, which is why the cross-Wurtz is a poor synthesis.
Two step mechanism of the Wurtz reaction and why mixed halides give three products Step one of the Wurtz reaction, an alkyl halide reacts with sodium to give an organosodium intermediate R Na. Step two, that intermediate attacks a second molecule of alkyl halide and displaces the halide, joining the two alkyl groups to give R R. If two different halides R X and R prime X are mixed, all three combinations form, R R, R R prime and R prime R prime, so the yield of the wanted product is poor and separation is hard. Wurtz reaction: two steps, and one big limitation STEP 1 Sodium makes an organosodium intermediate R–X + 2Na dry ether R–Na + NaX carbanion-like STEP 2 It attacks a second halide molecule R–Na + X–R R–R + NaX carbons doubled MIX TWO DIFFERENT HALIDES AND YOU GET ALL THREE R–X + R′–X R–R R–R′ R′–R′ unwanted the one you wanted unwanted Three products with similar boiling points, so separation is painful. Use Corey-House instead. ONLY EVEN-NUMBERED ALKANES two equal halves join, so C2, C4, C6 ... METHANE IS IMPOSSIBLE the smallest product is ethane
Fig 11The Wurtz reaction in two steps. Step 1 makes an organosodium, step 2 uses it as a nucleophile on a second . Because the joining is random, two different halides give three alkanes, and can never be made this way.

4.2 Frankland reaction

The same coupling idea with zinc in ethanol in place of sodium in ether.

4.3 Corey-House synthesis

BEYOND CORE SYLLABUS The Corey-House synthesis is not named in the NCERT Class 11 Hydrocarbons chapter or in the JEE Main syllabus list, but it appears in most coaching material and has featured in JEE Advanced-style questions on alkane synthesis. Read it for completeness; it is not a NEET requirement.
Corey House synthesis of butane from chloroethane in three steps Step one: chloroethane plus two lithium in dry ether gives ethyllithium and lithium chloride. Step two: two ethyllithium plus copper iodide gives lithium diethylcuprate, the Gilman reagent, plus lithium iodide. Step three: the cuprate reacts with chloroethane to give n-butane. Unlike the Wurtz reaction this gives a single unsymmetrical product. 1 CH₃CH₂Cl + 2Li → CH₃CH₂Li + LiCl dry ether, forms the alkyllithium 2 2CH₃CH₂Li + CuI → (CH₃CH₂)₂CuLi + LiI lithium dialkylcuprate, the Gilman reagent 3 (CH₃CH₂)₂CuLi + CH₃CH₂Cl → n-butane the cuprate delivers one alkyl group to the halide Why this beats the Wurtz reaction R₂CuLi couples with R'X to give R—R' only, with no R—R or R'—R' contamination It also works with aryl and vinyl halides, which the Wurtz reaction cannot touch
Fig 12Corey-House synthesis. Three steps, but unlike the Wurtz reaction it gives only the wanted and no symmetrical by-products.

4.4 Kolbe electrolytic synthesis

A concentrated aqueous solution of the sodium or potassium salt of a carboxylic acid is electrolysed. At the anode the carboxylate loses an electron, then loses , and the two alkyl radicals combine.

Kolbe electrolytic synthesis of an alkane from a potassium carboxylate An electrolysis cell containing concentrated aqueous potassium acetate. At the anode the carboxylate loses an electron, loses carbon dioxide and the two alkyl radicals combine to give ethane. At the cathode potassium hydroxide and hydrogen gas are produced. The alkane has twice the alkyl carbon count of the starting acid. conc. aqueous CH₃COOK Anode (+) Cathode (−) At the anode CH₃—CH₃ (the alkane) and 2CO₂ At the cathode 2KOH and H₂ 2CH₃COOK + 2H₂O → CH₃—CH₃ + 2CO₂ + 2KOH + H₂
Fig 13Kolbe electrolysis. The alkane forms at the anode, hydrogen at the cathode, and two alkyl groups combine so the carbon count doubles.
  1. at the anode
  3. , the alkane
Side products are always present: an ester from combining with , plus an alkene and a lower alkane from disproportionation of two alkyl radicals, for example .

4.5 Hydroboration coupling

BEYOND CORE SYLLABUS Diborane adds to an alkene to give a trialkylborane, which couples on treatment with at about to give a long-chain alkane. Useful background, but not part of the NCERT or JEE Main treatment of alkane preparation.

5Decarboxylation (One Carbon Fewer)

Sodium salts of carboxylic acids lose when heated with soda lime, a mixture of sodium hydroxide and calcium oxide. Calcium oxide is there to keep the mixture dry and make it easier to handle, not to react.

Decarboxylation of a sodium carboxylate with soda lime The sodium salt of a carboxylic acid heated with soda lime, a mixture of sodium hydroxide and calcium oxide, loses carbon dioxide to give a carbanion. The carbanion takes a proton to give the alkane. The alkane has one carbon fewer than the acid. Sodium formate is the exception and gives hydrogen gas instead. R C O O Na − + soda lime, Δ NaOH + CaO R − + CO₂ HOH R—H the carbanion takes a proton Carbon count drops by one: n carbons in the acid, n minus 1 in the alkane Exception: sodium formate HCOONa has no alkyl group, so it gives H₂ instead of R—H
Fig 14Decarboxylation. Losing costs one carbon, so an acid with carbons gives an alkane with .
Sodium formate, , has no alkyl group at all. Decarboxylating it gives hydrogen gas, not an alkane. This is asked directly.

6Methods That Give Methane Only

  • Hydrolysis of aluminium carbide:
  • Hydrolysis of beryllium carbide:
  • Berthelot synthesis: , and
Note the contrast with calcium carbide, , which hydrolyses to ethyne, not an alkane. Aluminium and beryllium carbides are methanides; calcium carbide is an acetylide.

7Choosing a Method in an Exam Question

Choosing a preparation method by the carbon count you need A number line marked n minus 1, n and 2n. Decarboxylation with soda lime gives n minus 1 carbons. All reduction methods keep the count at n. The Wurtz, Frankland, Kolbe and Corey House methods double it to 2n. Sodium acetate gives methane by decarboxylation but ethane by Kolbe electrolysis. n − 1 n 2n One carbon lost Soda lime decarboxylation Count unchanged Every reduction method Count doubled Wurtz, Frankland, Kolbe, Corey-House From CH₃COONa: soda lime gives CH₄, Kolbe gives CH₃CH₃ Same starting material, different method, different answer
Fig 15Exam questions usually tell you the target. Count the carbons first, and the method chooses itself.
Every alkane preparation reagent sorted by what it does to the carbon count A revision card sorting every reagent by carbon count. Same carbon count, hydrogen with nickel palladium or platinum, zinc with hydrochloric acid, lithium aluminium hydride for primary and secondary halides only, red phosphorus with hydroiodic acid, zinc amalgam with hydrochloric acid for the Clemmensen reduction, hydrazine with potassium hydroxide for the Wolff Kishner reduction, and a Grignard reagent with water. Carbon count increased, sodium in dry ether for the Wurtz reaction, zinc for the Frankland reaction, lithium dialkylcuprate for the Corey House synthesis and electrolysis of a potassium carboxylate for the Kolbe synthesis. Carbon count reduced, soda lime which is sodium hydroxide with calcium oxide on heating, which removes one carbon as carbon dioxide. One card, every reagent, sorted by carbon count CARBON COUNT UNCHANGED H2 / Ni, Pd or Pt alkene or alkyne → alkane Zn + dilute HCl R–X → R–H (nascent hydrogen) LiAlH4 1° and 2° R–X only; 3° gives an alkene red P + HI, Δ R–X or R–OH → R–H Zn–Hg + conc. HCl · Clemmensen C=O → CH2, acidic conditions N2H4 then KOH, Δ · Wolff-Kishner C=O → CH2, basic conditions RMgX + H2O · Grignard hydrolysis MORE CARBONS Na / dry ether Wurtz · carbons doubled Zn, Δ Frankland · carbons doubled R2CuLi + R′X Corey-House · R joins R′ electrolysis, RCOOK Kolbe · gives 2n−2 carbons ONE CARBON FEWER NaOH + CaO, Δ soda lime · decarboxylation RCOONa → R–H, one C lost as CO2
Fig 16The whole chapter on one revision card. In an exam, count the carbons in the target alkane against the carbons in the starting material first; that single comparison eliminates two of these three columns immediately.

Solved Examples

Solved Example 1
Prepare n-butane from chloroethane using the Corey-House synthesis.
Solution:

Step 1.

Step 2. , the Gilman reagent

Step 3.

Two carbons plus two carbons gives the four-carbon product, n-butane.

Solved Example 2
Prepare deuteropropane, , from isopropyl bromide.
Solution:

Step 1. Make the Grignard reagent:

Step 2. Quench with heavy water instead of ordinary water:

The Grignard carbon is , so it takes the deuteron exactly where the hydrogen would have gone. Using here would simply give propane.

Solved Example 3
Starting from , how would you obtain (i) methane and (ii) ethane?
Solution:

(i) Methane. Heat with soda lime. Decarboxylation removes one carbon.

(ii) Ethane. Electrolyse the concentrated aqueous solution. The two methyl radicals combine at the anode.

Same starting material, opposite effects on the carbon count.

Solved Example 4
Why can methane not be prepared by the Wurtz reaction?
Solution:

The Wurtz reaction joins two alkyl groups: . The smallest alkyl group is methyl, so the smallest possible product is , ethane, which has two carbons.

There is no way to arrive at a one-carbon product from a coupling reaction. More generally, the Wurtz reaction can only make alkanes with an even number of carbon atoms when both halides are the same.

Solved Example 5
What happens when 2-bromo-2-methylpropane is treated with (i) and (ii) in dry ether?
Solution:

(i) The substrate is a alkyl halide. is a strong base as well as a hydride source, so elimination beats substitution and the product is 2-methylpropene, , not the alkane.

(ii) Tertiary halides also fail in the Wurtz reaction for the same steric and basicity reasons, giving the alkene rather than 2,2,3,3-tetramethylbutane.

Both parts hinge on one idea: a halide with a strong base eliminates.

Solved Example 6
An alkane of formula is obtained by the Wurtz reaction, and the same alkane is also obtained by Kolbe electrolysis of a salt . Identify the halide used and the salt .
Solution:

For the Wurtz route, requires , so the halide is bromoethane, .

For the Kolbe route, the alkane is also with , so the carboxylate must be . That is the propanoate ion, so is sodium propanoate, .

Note that the Kolbe salt has three carbons, not two, because one carbon is lost as from each half.

Common Mistakes to Avoid

Watch out
  • Confusing with soda lime on a carboxylic acid. keeps all carbons and gives ; soda lime removes one and gives .
  • Forgetting that Kolbe needs a three-carbon salt for a four-carbon alkane. Each half loses one carbon as before coupling.
  • Trying to make methane by Wurtz or Kolbe. Both are coupling reactions, so the minimum product is ethane.
  • Using on a halide and expecting the alkane. You get the alkene.
  • Writing as a methane source. Calcium carbide gives ethyne. Aluminium and beryllium carbides give methane.
  • Quenching a Grignard reagent and forgetting the reagent is destroyed. Any protic solvent, including alcohol and ammonia, kills it instantly to the alkane. That is why Grignard reactions run in dry ether.
  • Using a cross-Wurtz in a synthesis answer. Three products means the examiner is testing whether you know to reach for Corey-House instead.

Frequently Asked Questions

What is the easiest way to remember all the preparation methods of alkanes?

Sort them by carbon count. Reductions of alkenes, alkynes, halides, alcohols and carbonyls keep the count at . Wurtz, Frankland, Kolbe and Corey-House double it to . Soda lime decarboxylation drops it to . Once you know the target alkane, the group of methods is fixed and you only have to recall the reagent.

Why is the Wurtz reaction not used for two different alkyl halides?

Two different halides give three alkanes: , and the wanted . Because the three have similar boiling points and similar polarity, separating them is difficult and the yield of the desired product is low. The Corey-House synthesis solves this by delivering only the unsymmetrical product.

Why can methane not be prepared by the Wurtz reaction?

The Wurtz reaction couples two alkyl groups, so the product always has at least two carbon atoms. The smallest alkyl group is methyl, and coupling two methyls gives ethane. For methane, use decarboxylation of sodium acetate with soda lime, or hydrolysis of aluminium or beryllium carbide.

What is the difference between Clemmensen and Wolff-Kishner reduction?

Both convert into . Clemmensen uses zinc amalgam with concentrated and therefore needs an acid-stable substrate. Wolff-Kishner uses hydrazine with dilute on heating and works in basic conditions, so it is the choice when the molecule cannot survive strong acid. They are complementary, which is why both are taught.

Why does give an alkene with tertiary alkyl halides?

supplies hydride, which is both a nucleophile and a strong base. A tertiary carbon is too hindered for backside attack, so instead of substituting, the hydride removes a beta hydrogen and an elimination occurs. The product is the alkene, following the Saytzeff preference for the more substituted double bond.

What is produced at the anode and cathode in Kolbe electrolysis?

The alkane and carbon dioxide are produced at the anode, where the carboxylate is oxidised, loses and the alkyl radicals combine. Hydrogen gas and the alkali metal hydroxide appear at the cathode. Side products include an ester and an alkene from radical disproportionation.

Why is calcium oxide added to sodium hydroxide in soda lime?

Calcium oxide keeps the mixture dry and less corrosive, and it makes the solid easier to handle and to heat in glass apparatus. It takes no chemical part in the decarboxylation itself; the sodium hydroxide does the work.

How do you get a deuterated alkane from an alkyl halide?

Convert the alkyl halide into a Grignard reagent with magnesium in dry ether, then quench it with heavy water, , instead of ordinary water. The carbon of the Grignard reagent is and picks up the deuteron, giving in place of .

Previous year questions on Preparation of Alkanes

2 questions from past papers, each with a step-by-step solution.

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