THC is the active component in marijuana, and ethanol is the alcohol in alcoholic beverages. Explain why drug screenings are able to detect the presence of THC but not ethanol weeks after these substances have been introduced into the body. CH3 OH CH;CH2-OH ethanol CH3 ČH3 (CH),CH3 tetrahydrocannabinol THC
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- THC is the active component in marijuana, and ethanol is the alcohol inalcoholic beverages. Explain why drug screenings are able to detect thepresence of THC but not ethanol weeks after these substances havebeen introduced into the body.Name the following ketones and aldehydes. When possible, give both a common name and an IUPAC name. ) CH3CH2CHBrCH2CH(CH3)CHODraw the structure of CH3CH(OH)CH2CH2CHO, where -CHO represents an aldehyde group and answer the following questions: 1. What is the name of this compound? 2. Does this compound contain a chiral carbon? If so, what number carbon is chiral? 3. Does this compound give a positive Tollen's test? If so, what would be the observations for this test? Draw the structure for the product of adding Tollen's reagent to the aldehyde? No molecular formulas. 4. What is oxidized? What is reduced? 5. This compound can form a cyclic hemiacetal, since it has an internal OH group and an aldehyde group. What size ring would be formed from this cyclic hemiacetal?
- Write the complete reaction of a. Oxidation of CH3CH2CH2CH2CH2CH2OH b. Oxidation of CH3CHOHCH2CH2CH2CH3 c. Dehydration of 2-methylcyclopentanol d. Dehydration of 2 moles of butanol e. Dehydration of 2-butanol f. Dehydration of 3-methyl-2-butanolComparing 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−.Arrange the following compounds in order of increasing boiling point.Explain. (CH3)2CHCH2OH (CH3)2CH(CH2)2CH2OH (CH3)2CH(CH2)3CH2OH
- What will happen when CH3NH3Cl is added to a CH3NH2 solution?) Which of the following compounds will be most soluble in ethanol (CH3CH2OH)? A) ethylene glycol (HOCH2CH2OH) B) hexane (CH3CH2CH2CH2CH2CH3) C) acetone (CH3COCH3)D) trimethylamine (N(CH3)3) E) A and D are soluble in ethanol.Maltose is a carbohydrate present in malt, the liquid obtained from barley and other grains. Although maltose has numerous functional groups, its reactions are explained by the same principles we have already encountered.a. Label the acetal and hemiacetal carbons.b. What products are formed when maltose is treated with each of these reagents: [1] H3O+; [2] CH3OH and HCl; [3] excess NaH, then excess CH3I?c. Draw the products formed when the compound formed in Reaction [3] of part (b) is treated with aqueous acid.The reactions in parts (b) and (c) are used to determine structural features of carbohydrates like maltose.
- Draw the products formed when CH3CH2C=C Na+ reacts with following compound. (CH3)2CHCH2CH2Cl1) identify the reactant/s in the chemical equation and circle it, also name its major functional group 2) identify the product/s in the chemical equation (circle it) and name its functional group 3) is the a reversible reaction or not? How do we knowEthyl butyrate, CH3CH2CH2CO2CH2CH3CH3CH2CH2CO2CH2CH3, is an artificial fruit flavor commonly used in the food industry for such flavors as orange and pineapple. Its fragrance and taste are often associated with fresh orange juice, and thus it is most commonly used as orange flavoring. It can be produced by the reaction of butanoic acid with ethanol in the presence of an acid catalyst (H+H+): CH3CH2CH2CO2H(l)+CH2CH3OH(l)H+⟶CH3CH2CH2CO2CH2CH3(l)+H2O(l) Part A Given 7.30 gg of butanoic acid and excess ethanol, how many grams of ethyl butyrate would be synthesized, assuming a complete 100%% yield? Express your answer in grams to three significant figures. Part B A chemist ran the reaction and obtained 5.95 gg of ethyl butyrate. What was the percent yield? Express your answer as a percent to three significant figures. Part C The chemist discovers a more efficient catalyst that can produce ethyl butyrate with a 78.0%% yield. How many grams would be produced from 7.30 gg of…