Question of 130
Q.Giving justification, categorise the following molecules/ions as nucleophile or electrophile: C6H5^-, BF3, C2H5O^-, (CH3)3N, Cl^+, CH3CO (charge/dot mark above the C not fully legible in the scan), NH2 (charge/bar mark above the N not fully legible), NO2 (charge/dot mark above the N not fully legible)
Rajasthan RbseRajasthan Board Senior Secondary Part-I Examination 2017Subjective· 3mImportance★★★★★
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Start your 14-day free trial to unlock the full solution →Species with a negative charge or an available lone pair (electron-rich) are nucleophiles; species with a positive charge or an incomplete octet (electron-deficient) are electrophiles.
A nucleophile ("nucleus-loving") is an electron-rich species that can donate an electron pair to an electron-deficient centre, forming a new bond -- typically an anion or a neutral species with a lone pair. An electrophile ("electron-loving") is an electron-deficient species that can accept an electron pair -- typically a cation or a neutral species with an incomplete octet.
Going through each species:
- C6H5^- (phenyl carbanion): negatively charged, carbon bears a lone pair -- NUCLEOPHILE.
- BF3: boron has only 6 electrons around it (incomplete octet), making it strongly electron-deficient -- ELECTROPHILE (classic Lewis acid).
- C2H5O^- (ethoxide ion): negatively charged oxygen with lone pairs -- NUCLEOPHILE.
- (CH3)3N (trimethylamine): neutral, but nitrogen carries a lone pair available for donation -- NUCLEOPHILE.
- Cl^+ (chloronium/chlorine cation): positively charged, electron-deficient -- ELECTROPHILE.
- CH3CO (species with a charge mark above the carbon that was not fully legible in the scan): read here as the acylium cation, CH3CO^+, i.e., CH3-C(triple bond)O^+ (or CH3-C=O)^+, positively charged and electron-deficient at carbon -- ELECTROPHILE. (Honest note: if the mark is instead meant to indicate a radical/dot rather than a full positive charge, the species would be a neutral acyl radical rather than a clean nucleophile/electrophile; the cationic reading is used here as the standard textbook version of this question.) …
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