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Question: What would be the compound D? In the following reaction: \[C{H_3} - C{H_2} - Br\xrightarrow[\Delt...

What would be the compound D? In the following reaction:
CH3−CH2−Br→ΔAlc.KOHB→HBrC→Na−etherDC{H_3} - C{H_2} - Br\xrightarrow[\Delta ]{{Alc.KOH}}B\xrightarrow{{HBr}}C\xrightarrow{{Na - ether}}D ?
A) Ethane
B) Propane
C) n-butane
D) n-heptane

Explanation

Solution

As we know that alkyl halides undergoes dehydrohalogenation when treated with alcoholic potassium ethoxide which will result in elimination reactions and hydrogen bromide as a reagent acts by addition of one halide into the compound and finally sodium/ dry ether undergoes breakage of double bonds.

Complete step by step solution:
As we know that alkyl halides when treated with alcoholic potassium ethoxide, it undergoes dehydrohalogenation which results into the formation of an alkene by undergoing an elimination reaction. Similarly, ethyl bromide is treated with alcoholic potassium ethoxide and forms ethylene. We can show this reaction with the help of equation:
CH3−CH2−Br→ΔAlc.KOHH2C=CH2C{H_3} - C{H_2} - Br\xrightarrow[\Delta ]{{Alc.KOH}}{H_2}C = C{H_2}
Ethylene will further under hydrohalogenation where double bond is broken down by the attack of bromine on the double bond and addition of a hydrogen atom to the other carbon resulting again in the formation of ethyl bromide will take place. We can show this reaction by below equation as:
H2C=CH2→HBrCH3−CH2−Br{H_2}C = C{H_2}\xrightarrow{{HBr}}C{H_3} - C{H_2} - Br
Then, the haloalkane so formed will react with metals in the presence of dry ether forming organometallic compounds which we normally call the Wurtz reaction and two molecules of alkyl halide will result in the formation of a single bonded linear alkane molecule. So, two molecules of ethyl bromide will react with sodium/dry ether and result in the formation of four carbon compounds which will be butane. We can show this reaction as:
CH3−CH2−Br→Na−etherCH3−CH2−CH2−CH3C{H_3} - C{H_2} - Br\xrightarrow{{Na - ether}}C{H_3} - C{H_2} - C{H_2} - C{H_3}
Therefore, we can show the complete reaction as:
CH3−CH2−Br→ΔAlc.KOHH2C=CH2→HBrCH3−CH2−Br→Na−etherCH3−CH2−CH2−CH3C{H_3} - C{H_2} - Br\xrightarrow[\Delta ]{{Alc.KOH}}{H_2}C = C{H_2}\xrightarrow{{HBr}}C{H_3} - C{H_2} - Br\xrightarrow{{Na - ether}}C{H_3} - C{H_2} - C{H_2} - C{H_3}

Therefore, from the above explanation we can say that the correct answer is (C).

Note: Remember that alcoholic KOHKOH is potassium ethoxide which helps in forming ethoxide ion and aqueous KOHKOH is potassium hydroxide which help in the formation of hydroxide ion upon dissociation. Alcoholic potassium ethoxide generally undergo elimination reactions whereas aqueous potassium hydroxide undergo substitution reactions.