Indian Navy SSR Chemistry — practice by topic

**Test.** For cis–trans isomerism each sp²-carbon of the C=C bond must carry two *different* groups. If either carbon has two of the same substituent, the two configurations are identical. - **A. 2-methylpropene** $(CH_3)_2C=CH_2$ — left carbon has two CH₃ groups → no. - **B. Propene** $CH_3CH=CH_2$ — the CH₂ end has two H → no. - **C. But-2-ene** $CH_3CH=CHCH_3$ — each sp²-C carries one CH₃ and one H → **yes**, gives cis (Z) and trans (E). - **D. Pent-1-ene** $CH_2=CHCH_2CH_2CH_3$ — terminal C has two H → no. The rule of thumb: terminal alkenes never show cis–trans internal alkenes do *only if* the two substituents on each end differ.

**Cannizzaro reaction** is the disproportionation of an aldehyde with **no $\alpha$-hydrogen** under concentrated alkali. One molecule of the aldehyde is oxidised, one is reduced. For benzaldehyde + NaOH (conc.): - One PhCHO is *oxidised* to PhCOO⁻Na⁺ (**sodium benzoate**). - One PhCHO is *reduced* to PhCH₂OH (**benzyl alcohol**). So the product mix is option C. Acidifying the workup converts the sodium benzoate to benzoic acid, but the raw Cannizzaro products are as written. **Why option A is wrong on its own.** Plain oxidation of benzaldehyde with stronger oxidants gives benzoic acid alone that's *not* Cannizzaro — Cannizzaro requires the simultaneous reduction half.

P in $\mathrm{PCl_5}$ has five bonded chlorines and no lone pair. That gives five electron domains. The hybridisation needed is $sp^3d$ (one $s$ + three $p$ + one $d$ orbital = five hybrid orbitals). The molecular geometry is trigonal bipyramidal. $sp^3d^2$ would give six domains and an octahedral geometry, which is what you see in $\mathrm{SF_6}$, not $\mathrm{PCl_5}$.