When $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHCl}_2$ is treated with $\mathrm{NaNH}_2$, the product formed is:

1. $\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2$
2. $\mathrm{CH}_3-\mathrm{C} \equiv \mathrm{CH}$
3. 
   
4. 
   

Asked in: NEET 2002

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