In this compound groups preferential order is $$ - Cl > - C{H_3}\,{\text{and}}\, - I > - C{H_2} - C{H_3}$$
Hence, more preferential order containing groups are attached at opposite sides. So, it is $$E$$ $$(trans)$$  - isomer. Thus, its name is $$trans-2-chloro-3-iodo-2-pentene.$$

Thus, $${2^ \circ }$$ carbocation is formed.

No explanation is given for this question. Let's discuss the answer together.

$$ - Cl{\text{ - }}atom$$   shows $$+R$$ - effect that $$o/p$$ -directive influence but deactivate the benzene ring. While $$ - OH, - C{H_3}$$   group also shows $$o/p$$ -influence but activate the benzene ring. But in these $$-OH$$  group activates more than $$ - C{H_3}.$$
Hence, order of electrophilic substitution is

Steam distillation is used to separate substances which are steam volatile and are immiscible with water.

Because of the presence of $$7$$ $$p$$ - orbitals and 7 unpaired electrons, it is aromatic in nature as these unpaired electrons delocalise in $$p$$ - orbitals.

The enol form of acetone is
  with $${D_2}O$$  it gives

The numbering of $$C$$ - atom starts from $$\mathop C\limits^ * $$  or $$\mathop C\limits^{ * * } .$$  But numbering from $$\mathop C\limits^ * $$  give minimum locant $$(2)$$  to $$Br$$  which is correct.

In IUPAC system of nomenclature, preference is given to multiple bond than halogen substituent, so the correct name of $$Br - \mathop {\mathop {C{H_2}}\limits^3 - \mathop {CH}\limits^2 = \mathop {C{H_2}}\limits^1 }\limits_{3{\text{ }} - bromoprop - 1 - ene} $$      is Priority is given to double bond than halogen.

Priority of functional groups is
Hence, the IUPAC name is 4-formyl-2-oxocyclohexane carboxylic acid.