See image — IUPAC and Nomenclature Chemistry Question

💡 Solution & Explanation

To identify compounds that are BOTH an aldehyde AND an ether, we need to find structures containing: (1) an aldehyde functional group: -CHO (a carbonyl carbon bonded to at least one hydrogen), and (2) an ether functional group: C-O-C linkage (oxygen bonded to two carbon atoms, not part of a carbonyl). Analysis of each structure: (i) This structure shows a carbon chain with an ether linkage (C-O-C, specifically ethyl ether) AND a -CH=O aldehyde group hanging off the branched carbon. It contains both an ether oxygen (C-O-C) and an aldehyde (-CHO). Therefore, (i) qualifies. (ii) This structure shows a linear chain with a ketone (C=O in the middle, flanked by two carbons) and an ether oxygen at one end. The carbonyl is a ketone, NOT an aldehyde (no H on the carbonyl carbon). So (ii) does NOT qualify as an aldehyde. (iii) This appears to be a simple molecule showing only an epoxide or a carbonyl group with no aldehyde functionality and possibly just an epoxide (cyclic ether). There is no aldehyde group visible. So (iii) does NOT qualify. (iv) This structure shows a four-membered ring containing an oxygen (oxetane = cyclic ether, C-O-C linkage) AND an aldehyde group (-CHO, shown with the C=O and H explicitly). It contains both an ether (the oxetane ring oxygen) and an aldehyde (-CHO). Therefore, (iv) qualifies. (v) This structure shows a terminal aldehyde on one end and an ester group (-C(=O)OCH3) on the other end. An ester is NOT an ether (the oxygen is adjacent to a carbonyl). The OCH3 in an ester context does not count as a simple ether. So (v) does NOT qualify as containing an ether functional group in the classic sense. Thus, only (i) and (iv) contain both an aldehyde and an ether functional group. Why other options fail: - (a) (i) only: misses (iv) which also has both groups. - (c) (ii) and (v): (ii) has a ketone not aldehyde; (v) has an ester not a simple ether. - (d) (iii) and (iv): (iii) lacks an aldehyde group. Therefore, the correct answer is B.