Explanation
The products formed in the reaction is(are):
Ethyl formate is written as $$HCOO{C_2}{H_5}$$ . The reaction is as follows
$$HCOO{C_2}{H_5} + RMgX \to RCHO + Mg(O{C_2}{H_5})X$$
Hence the correct answer is option A
4 3 2 1
$$CH_{3}-\underset {\underset{\displaystyle {CH_{3}} }{\displaystyle |}}{C}H–CH_{2}–CH_{2}Br$$
Its IUPAC name is 1-bromo-3-methylbutane.
The correct option is B.
As refer to the above image,
TAM- Anti addition to Trans results in meso compound.
$$(a)$$ As refer to the IMAGE $$01\rightarrow$$ Due to geometrical Isomerism, it has structure at $$C-2$$
$$(b)$$ As refer to the IMAGE $$02\rightarrow$$ Possible structure has $$4$$ isomers and $$5$$ stereoisomers.
$$(c)$$ As refer to the IMAGE $$03\rightarrow$$ Only at terminal carbons same for any substitution.
$$(d)$$ As refer to the IMAGE $$04\rightarrow$$ Structure are $$3$$ stereoisomers.
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