Product (C) in the above reaction is:

1. 

2.

3. 

4.

The product of the given reaction is :

; Product (B) of this reaction is :

1.		2.
3.		4.

Cyclopentadiene is much more acidic than cyclopentane, because:

 The compounds among the following that can be generated by Friedel craft acylation is/are -

I.
II.
III.
IV.

     
The product B in the above-mentioned reaction is:

1.		2.
3.		4.

The major product obtained in the given reaction is:

1.		2.
3.		4.

$\underset{\left(2\right)<mspace\; width="1em"></mspace>{\mathrm{H}}_3{\mathrm{O}}^{\oplus }}{\overset{\left(1\right)<mspace\; width="1em"></mspace>{\mathrm{KMnO}}_4,<mspace\; width="1em"></mspace>{\mathrm{HO}}^{-},<mspace\; width="1em"></mspace>∆}{\to }}$

The labelled carbon goes with:

$1.<mspace\; width="1em"></mspace>\mathrm{Ph}-\stackrel{14}{\mathrm{C}}{\mathrm{O}}_2\mathrm{H}<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>$
$2.<mspace\; width="1em"></mspace>\stackrel{14}{\mathrm{C}}{\mathrm{O}}_2<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>$
$3.<mspace\; width="1em"></mspace>\mathrm{Ph}-\stackrel{14}{\mathrm{C}}{\mathrm{H}}_2-{\mathrm{CO}}_2\mathrm{H}<mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace><mspace\; width="1em"></mspace>$
$4.<mspace\; width="1em"></mspace>\stackrel{14}{\mathrm{C}}{\mathrm{H}}_4$

The deactivating but ortho-para directing benzene ring substituent among the following is-

1. -N=O

2. -OCH₃

3. -COCH₃

4. -NO₂

Which of the following organic chlorides will not give a Friedel-Craft alkylation product when heated with benzene and AlCl₃/FeCl₃?