The major product of the following reaction is
CH3CH2CH(Br)-CH2(Br) - (i) KOH alc. (ii) NaNH2 in li

Q: Why can't we just use excess Alcoholic KOH to get the alkyne directly?

Alcoholic KOH is strong enough to remove the first HBr molecule, forming a vinyl halide. However, in a vinyl halide, the lone pair on the halogen is in resonance with the double bond, giving the C-X bond partial double bond character. Alcoholic KOH is not strong enough to break this stabilized bond.

Q: What is the difference between using Alcoholic KOH and Aqueous KOH?

This is a crucial distinction! Alcoholic KOH acts as a strong base and favors elimination (forming alkenes/alkynes). Aqueous KOH acts as a nucleophile and favors substitution (replacing halogens with OH groups to form alcohols/diols).

Q: Why does the reaction form a sodium alkynide salt in the intermediate step?

The product of the second elimination is a terminal alkyne. The hydrogen on a terminal alkyne is attached to an sp-hybridized carbon, making it quite acidic. Since sodamide ($NaNH_2$) is a very strong base, an acid-base reaction occurs immediately, forming the salt.

Q: What if the starting material was a geminal dihalide instead of a vicinal dihalide?

The reaction sequence would be exactly the same! A geminal dihalide (both halogens on the same carbon) would also undergo double dehydrohalogenation with these reagents to yield an alkyne.

Q: Why is the hydrogen on a terminal alkyne acidic?

The carbon atom in a terminal alkyne is sp-hybridized, meaning it has 50% s-character. The s-orbital is closer to the nucleus, making the sp-carbon highly electronegative. It pulls the electron density of the C-H bond towards itself, allowing the hydrogen to leave easily as an $H^+$ ion.

Question

The major product of the following reaction is

Options:

(A) $	ext{CH}_3	ext{CH}_2	ext{CH}(	ext{NH}_2)	ext{CH}_2	ext{NH}_2$

(B) $	ext{CH}_3	ext{CH}=	ext{CHCH}_2	ext{NH}_2$

(C) $	ext{CH}_3	ext{CH}=	ext{C}=	ext{CH}_2$

(D) $	ext{CH}_3	ext{CH}_2	ext{C}\equiv	ext{CH}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are looking at a molecule of **1,2-dibromobutane**. This is a classic example of a vicinal dihalide, where two halogen atoms are attached to adjacent carbon atoms.

Our goal is to transform this saturated molecule into an alkyne. To do this, we need to perform a double dehydrohalogenation—essentially ripping away two molecules of hydrogen bromide ($HBr$). The reagents provided are **alcoholic KOH** followed by **$NaNH_2$ in liquid ammonia**. Let's break down why this specific sequence is necessary.

### The First Elimination: Alcoholic KOH

Alcoholic KOH is a strong base, perfectly suited for our first step. When it attacks the vicinal dihalide, it initiates an $E2$ elimination reaction. It abstracts a proton from one carbon while the bromine atom leaves from the adjacent carbon.

$$CH_3CH_2CH(Br)CH_2Br \xrightarrow{Alc. KOH} CH_3CH_2C(Br)=CH_2$$

This single elimination yields an intermediate known as a **vinyl halide** (specifically, 2-bromo-1-butene).

### The Resonance Trap

Now, you might wonder, why can't we just use more alcoholic KOH to remove the second $HBr$ molecule? Here is the catch: vinyl halides are notoriously unreactive towards normal bases.

The lone pair of electrons on the bromine atom is in resonance with the newly formed carbon-carbon double bond. This delocalization gives the carbon-bromine bond a **partial double bond character**, making it significantly stronger and much harder to break. Alcoholic KOH simply isn't powerful enough to force the second elimination.

### The Master Equation: Enter Sodamide

To overcome this resonance stabilization, we must bring in the heavy artillery: **Sodamide ($NaNH_2$) in liquid ammonia**. Sodamide is an exceptionally strong base, far stronger than alcoholic KOH.

$$CH_3CH_2C(Br)=CH_2 \xrightarrow{NaNH_2} CH_3CH_2C\equiv CH$$

It forcefully abstracts the second proton, driving the elimination of the remaining bromine atom and successfully forging the carbon-carbon triple bond. We have now formed **1-butyne**.

### The Final Calculation: Acidic Workup

Don't make a silly mistake and stop here! The alkyne we just formed is a **terminal alkyne**. The hydrogen atom attached to the $sp$-hybridized carbon is highly acidic. Because we are operating in a strongly basic medium (sodamide is still present), an acid-base reaction occurs instantaneously.

$$CH_3CH_2C\equiv CH + NaNH_2 \rightarrow CH_3CH_2C\equiv C^-Na^+ + NH_3$$

The strong base abstracts the acidic proton, leaving us with a sodium alkynide salt. To isolate our desired neutral alkyne, we must perform a mild acidic workup (adding $H^+$) to re-protonate the alkynide ion.

$$CH_3CH_2C\equiv C^-Na^+ \xrightarrow{H^+} CH_3CH_2C\equiv CH$$

And there we have it! The final major product is **1-butyne**, making option (d) the correct answer. Always remember to respect the strength of your bases and the acidity of your products!

Animated Solution for Chemistry - Organic Chemistry: The major product of the following reaction is

Select Answer:

Visualized Solution

\text{Vicinal Dihalide}

\text{First Dehydrohalogenation}

\text{Formation of Vinyl Halide}

\text{The Resonance Trap}

\text{Second Dehydrohalogenation}

\text{Acidic Terminal Alkyne}

\text{Final Workup}

\text{The Way Forward}

The Sigma Insight: Hydrocarbons

Analyzing the Setup

The First Elimination

The Resonance Trap

The Master Equation

The Final Calculation

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