An unsaturated hydrocarbon X on ozonolysis gives A. Compound A when warmed with ammonical silver nitrate forms a bright silver mirror along the sides of the test tube. The unsaturated hydrocarbon X is :

Hydrocarbons PYQ — JEE Main 2021 · 16 Mar · Shift 2 · Q3 with Solution

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Correct answer: A

Interpret the clue from ozonolysis

Ozonolysis of an unsaturated hydrocarbon cleaves the multiple bond and gives carbonyl-type products.
Use the test with ammoniacal silver nitrate

Compound $A$ gives a bright silver mirror with ammoniacal silver nitrate, i.e. Tollens' reagent.

This test is given by aldehydes (and formic acid), so $A$ must be an aldehyde.
Check which alkyne on ozonolysis can give an aldehyde-type product

For alkynes:
- Internal alkynes on ozonolysis give carboxylic acids.
- Terminal alkynes can give a product corresponding to formyl/carboxyl end, and the key standard JEE clue is that only the terminal alkyne here can lead to a product giving Tollens' test.
Evaluate the given options

Option A: $HC \equiv C-CH_2-CH_3$

This is a terminal alkyne (1-butyne). It is the suitable unsaturated hydrocarbon among the options.

Option C: $CH_3-C \equiv C-CH_3$

This is an internal alkyne (2-butyne). On ozonolysis, internal alkynes give carboxylic acids, which do not give Tollens' silver mirror test.
Conclusion

Therefore, the unsaturated hydrocarbon $X$ is: $HC \equiv C-CH_2-CH_3$

So, Option A is correct.

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Correct answer: A

Interpret the clue from ozonolysis

Ozonolysis of an unsaturated hydrocarbon cleaves the multiple bond and gives carbonyl-type products.
Use the test with ammoniacal silver nitrate

Compound $A$ gives a bright silver mirror with ammoniacal silver nitrate, i.e. Tollens' reagent.

This test is given by aldehydes (and formic acid), so $A$ must be an aldehyde.
Check which alkyne on ozonolysis can give an aldehyde-type product

For alkynes:
- Internal alkynes on ozonolysis give carboxylic acids.
- Terminal alkynes can give a product corresponding to formyl/carboxyl end, and the key standard JEE clue is that only the terminal alkyne here can lead to a product giving Tollens' test.
Evaluate the given options

Option A: $HC \equiv C-CH_2-CH_3$

This is a terminal alkyne (1-butyne). It is the suitable unsaturated hydrocarbon among the options.

Option C: $CH_3-C \equiv C-CH_3$

This is an internal alkyne (2-butyne). On ozonolysis, internal alkynes give carboxylic acids, which do not give Tollens' silver mirror test.
Conclusion

Therefore, the unsaturated hydrocarbon $X$ is: $HC \equiv C-CH_2-CH_3$

So, Option A is correct.