Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

Iso-propyl bromide on Wurtz reaction gives:
(A) Hexane
(B) Propane
(C) 2,3-dimethyl butane
(D) Neo-hexane

seo-qna
Last updated date: 13th Jul 2024
Total views: 357.3k
Views today: 8.57k
Answer
VerifiedVerified
357.3k+ views
Hint:
The structure of isopropyl group is
seo images

Wurtz reaction gives Sodium halide as a side product in the reaction.

Complete step by step answer:

Let’s first see what happens in the Wurtz reaction.
In Wurtz reaction, two alkyl halides couple in presence of sodium metal and give alkane and sodium halide in ether solvent. The reaction can be written as,
\[2R - X\xrightarrow[{Ether}]{{Na}}R - R + NaX\]

Now let’s see the structure of isopropyl bromide.
seo images

So, if isopropyl bromide will be allowed to react with Sodium metal in ether solvent then, sodium metal will act as an reducing agent and will form a salt sodium bromide. Then the two isopropyl groups will combine and will give an alkane as a final product. So, the reaction can be written as below.
seo images

So, we can conclude that 2,3-dimethylbutane is the final product of the reaction.

Therefore the correct answer is (C) 2,3-dimethyl butane.



Additional Information:
See some of the other similar looking reactions as Wurtz reaction to avoid confusion.
Fitting reaction:
\[2Ar - X\xrightarrow[{Ether}]{{Na}}Ar - Ar + NaX\]
Here we can see that two aromatic halides react with sodium metal in ether to give coupled product and sodium halides as a side product.
Wurtz-Fittig reaction:
\[R - X + Ar - X\xrightarrow[{Ether}]{{Na}}Ar - R + NaX\]
Here we can see that one alkyl halide and one aryl halides are used as a starting material. They combine in presence of sodium metal and ether as solvent to give aromatic alkanes and sodium halide.


Note:
Do not get confused by two other same looking reactions like Wurtz reaction, which is Fittig reaction and Wurtz –Fittig reaction. Remember that in the isopropyl group, the substituent group will get bonded with the \[{2^{nd}}\] carbon of the group, so do not get confused with the n-propyl group.