In the following sequence of reactions the products D is
`C-=CH overset(HBr)toAoverset(HBr)toBoverset(alcKOH)toCoverset(NaNH_(2))toD`. D is

Options

A. Ethanol
B. Ethyne
C. Ethanal
D. Ethene

Answer

To determine the final product D in the given sequence of reactions, we will analyze each step carefully.

1. **Starting Material**: The initial reactant is acetylene (HC≡CH).

2. **First Reaction (with HBr)**:
- Acetylene reacts with HBr to form vinyl bromide (CH2=CHBr).
- The reaction can be represented as:
\[ HC \equiv CH + HBr \rightarrow CH_2=CHBr \quad (A) \]

3. **Second Reaction (with HBr again)**:
- The vinyl bromide (A) reacts with HBr again. According to Markovnikov's rule, the bromine (Br) will add to the carbon with fewer hydrogen atoms.
- The product formed will be 1-bromopropane (CH3-CHBr-CH2).
- The reaction can be represented as:
\[ CH_2=CHBr + HBr \rightarrow CH_3-CHBr-CH_2 \quad (B) \]

4. **Third Reaction (with alcoholic KOH)**:
- The product B undergoes beta elimination in the presence of alcoholic KOH. This results in the elimination of HBr, forming another alkene.
- The product formed will be 1-bromopropene (CH2=CHBr).
- The reaction can be represented as:
\[ CH_3-CHBr-CH_2 + alc.KOH \rightarrow CH_2=CHBr \quad (C) \]

5. **Fourth Reaction (with NaNH2)**:
- The alkene (C) reacts with sodium amide (NaNH2). The bromine atom is replaced by a hydrogen atom from the adjacent carbon, resulting in the formation of a terminal alkyne.
- The product formed will be propyne (HC≡C-CH3).
- The reaction can be represented as:
\[ CH_2=CHBr + NaNH_2 \rightarrow HC \equiv C-CH_3 \quad (D) \]

Thus, the final product D is propyne (HC≡C-CH3).

### Summary of Steps:
1. Acetylene reacts with HBr to form vinyl bromide (A).
2. Vinyl bromide reacts with HBr to form 1-bromopropane (B).
3. 1-bromopropane undergoes beta elimination with alcoholic KOH to form 1-bromopropene (C).
4. 1-bromopropene reacts with NaNH2 to form propyne (D).