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- [10] 10] Q1. Do the following conversions as directed. Write the complete reaction equation. (a) Benzaldehyde to Benzoic Acid. (b) Propanal to 1-propanal (c) Cyclohexanone to Cyclohexanol (d) Acetaldehyde to Ethyl alcohol (e) Acetone to Iso-propyl alcohole) Acetaminophen Exp: -NH2 Но2-pentanone +H2O
- 17-74 Glucose, C6H12O6, contains an aldehyde group but exists predominantly in the form of the cyclic hemiacetal shown here. We will discuss this cyclic form of glucose in Chapter 20. A cyclic hemiacetal is formed when the —OH group of one carbon bonds to the carbonyl group of another carbon. (a) Which carbon in glucose provides the —OH group and which provides the —CHO group? (b) Draw the alternative chair confirmations of D-glucose and state which of the two is the more stable.CH,- CH = 0 aldehyde CH, О— С — С — Н CH3 NH,–C=O | - CH3 CH, CH, — CH, —С — ОНplase solve the attached picture
- This molecule is a(n): || НаС — СН С — ОН CH- C - OH O carboxylic acid O alkyne O ether O alcoholName the following compound. O 2 -{N(CH₂)₂ Not aromatic O cyclohexanoyldimethylamine O dimethylaminocyclohexylketone N,N-dimethylcyclohexane lactam O dimethylaminooxocyclohexane O N,N-dimethylcarboxamidesolve the following
- Draw the products and balance the equation.Comparing Two Different Methods of Hydration of an Alkene Draw the product formed when CH3CH2CH2CH2CH=CH2 is treated with either (a) H2O, H2SO4; or (b) BH3 followed by H2O2, HO−.This molecule is a(n): HC = CH – CH3 | НаС — сн — сHз O amine O alkene O alkane O ether