Q.Which of the following reaction will give ionic organic product on reaction ?
A. $CH_3-CH_2-OH + Na \rightarrow$
B. $CH_3-CH_2-OH + SOCl_2 \rightarrow$
C. $CH_3-CH_2-OH + PCl_5 \rightarrow$
D. $CH_3-CH_2-OH + H_2SO_4 \rightarrow$

Step 1. Consider what each reagent does to ethanol. A. Sodium metal reacts with the weakly acidic O-H of ethanol: 2CH3-CH2-OH + 2Na -> 2CH3-CH2-O(-)Na(+) + H2 (section 11.4.4). The product, sodium ethoxide, is an ionic salt -- an alkoxide anion paired with a Na+ cation. B. SOCl2 converts the alcohol to ethyl chloride (a standard, covalent, molecular alkyl-halide-forming reaction, analogous to the PCl5/PX3 chemistry of section 11.4.4). C. PCl5 likewise converts the alcohol's C-O bond to give ethyl chloride, again a covalent molecular product. D. Concentrated H2SO4 with an alcohol typically drives dehydration to an alkene (section 11.4.4) or, at lower temperature, ether formation (section 11.5.1) -- again molecular, covalent products, not an isolable ionic organic species.

Step 2. Compare. Only option A produces a true ionic compound (the organic ethoxide anion, isolable as its sodium salt); B, C and D all give neutral, covalent organic products (alkyl chlorides or dehydration/ether products).

Checking which of the four reagents reacts at the O-H bond to break off H+ and leave an isolable ionic (alkoxide) organic species, versus reagents that give ordinary covalent substitution/elimination products

• Assuming PCl5 or SOCl2 give an 'ionic' product because a chloride ion is momentarily involved mechanistically; the actual isolated ORGANIC product in both cases is a neutral, covalent alkyl chloride.
• Overlooking that reaction with active metal Na is precisely the diagnostic alcohol test described in section 11.4.4 for exactly this reason -- it is the one reaction that liberates H2 and leaves an ionic alkoxide salt.