Question

1,1-Dibromopentane on treatment wtih sodamide followed by reaction with ${\mathrm{H}}_3{\mathrm{O}}^{+}$ gives (1) 2 -pentyne (2) 3 -mcthyl butyne-1 (3) 1-pentyne (4) $n$ -propyl cthyne

Short answer

Option (3) 1-pentyne

Step-by-step solution

- Identify the starting compound

The given compound is 1,1-Dibromopentane. Its structure can be written as CH3-CH2-CH2-CH2-CBr2.

- React with sodamide

When 1,1-Dibromopentane is treated with sodamide (NaNH2), a double dehydrohalogenation occurs. This will eliminate two bromine atoms and two hydrogen atoms to form a triple bond. The reaction is as follows: CH3-CH2-CH2-CH2-CBr2 + 2 NaNH2 → CH3-CH2-CH2-CH2-C≡CH

- Acidic workup

The next step involves the reaction of the intermediate alkyne with water in the presence of acid (H3O+), which essentially completes the pathway. However, in this case, the triple bond remains unaffected: CH3-CH2-CH2-CH2-C≡CH + H3O+ → CH3-CH2-CH2-CH2-C≡CH

- Identify the product

The resulting product after the final step is 1-pentyne. Therefore, the correct answer is option (3).

Key concepts

1,1-Dibromopentane

The compound 1,1-Dibromopentane is an organic molecule consisting of a five-carbon chain with two bromine atoms attached to the first carbon. Its molecular structure is: CH3-CH2-CH2-CH2-CBr2. This configuration is crucial because the presence of two bromine atoms on the same carbon makes it a suitable candidate for transformation through dehydrohalogenation, a process that will be detailed later.
In the context of the given exercise, understanding the structure helps in predicting the course of the chemical reaction.

Sodamide Reaction

Sodamide (NaNH2) is a strong base commonly used in organic chemistry to deprotonate compounds and initiate elimination reactions. When 1,1-Dibromopentane reacts with sodamide, a double dehydrohalogenation occurs, removing two bromine atoms and two hydrogen atoms. This process results in the formation of a triple bond, effectively converting a single-bonded carbon structure to an alkyne:
CH3-CH2-CH2-CH2-CBr2 + 2 NaNH2 → CH3-CH2-CH2-CH2-C≡CH + 2 NaBr + 2 NH3.
The byproducts of this reaction are sodium bromide (NaBr) and ammonia (NH3).

Triple Bond Formation

The formation of a triple bond is central to the transformation in this reaction. Through double dehydrohalogenation, the removal of two halogen atoms (bromine) and two hydrogen atoms allows the creation of a carbon-carbon triple bond. This results in an alkyne structure, specifically 1-pentyne (CH3-CH2-CH2-CH2-C≡CH).
Alkynes are characterized by having a triple bond between two carbon atoms, making them more reactive than their alkene and alkane counterparts.
Introducing a triple bond significantly alters the physical and chemical properties of the molecule, making it a valuable intermediate in various organic synthesis reactions.

Dehydrohalogenation

Dehydrohalogenation is a chemical reaction that involves the elimination of a hydrogen halide (HX) from an organic molecule. In the case of 1,1-Dibromopentane, the molecule undergoes double dehydrohalogenation when treated with a strong base like sodamide:
CH3-CH2-CH2-CH2-CBr2 + 2 NaNH2 → CH3-CH2-CH2-CH2-C≡CH.
In this process, each dehydrohalogenation removes one hydrogen (H) and one bromine (Br) atom, eventually leading to the formation of a triple bond between the carbon atoms. This mechanism is crucial in organic synthesis, especially for creating unsaturated compounds like alkynes.