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(a)
Interpretation:
The compound that has higher solubility in water among the given pair has to be identified.
Concept Introduction:
Carbonyl groups are the one which contain a double bond between carbon and oxygen atom.
Aldehydes contain a carbonyl group that contains a hydrogen atom and a carbon atom bonded to it. Aldehyde that has one and two carbon atoms are gas at room temperature. The physical state of aldehyde that contains three carbon atoms to eleven carbon atoms that are not branched is liquid at room temperature. Aldehydes that contain more than eleven carbon atoms are solid at room temperature.
Ketones contain a carbonyl group that contains two carbon atoms bonded to it. For a compound to be ketone, a minimum of three carbon atom is required. Ketones that contain three carbon atoms to eight carbon atoms which have the carbonyl group at the second carbon atom are liquid at room temperature.
Solubility of aldehydes and ketones depend upon the length of the carbon chain. Those contain less than six carbon atoms are soluble in both water and organic solvents. Aldehydes and ketones that contain six or more carbon atoms are not soluble in water but soluble in organic solvents only. Solubility of smaller ketones and aldehydes are result of the hydrogen bond formation between the lone pairs of oxygen atom and the hydrogen atom in the water molecule.
(b)
Interpretation:
The compound that has higher solubility in water among the given pair has to be identified.
Concept Introduction:
Carbonyl groups are the one which contain a double bond between carbon and oxygen atom. Aldehydes and ketones possess this carbonyl functional group in it. The structural representation of a carbonyl group can be given as shown below,
Aldehydes contain a carbonyl group that contains a hydrogen atom and a carbon atom bonded to it. Aldehyde that has one and two carbon atoms are gas at room temperature. The physical state of aldehyde that contains three carbon atoms to eleven carbon atoms that are not branched is liquid at room temperature. Aldehydes that contain more than eleven carbon atoms are solid at room temperature.
Ketones contain a carbonyl group that contains two carbon atoms bonded to it. For a compound to be ketone, a minimum of three carbon atom is required. Ketones that contain three carbon atoms to eight carbon atoms which have the carbonyl group at the second carbon atom are liquid at room temperature.
Solubility of aldehydes and ketones depend upon the length of the carbon chain. Those contain less than six carbon atoms are soluble in both water and organic solvents. Aldehydes and ketones that contain six or more carbon atoms are not soluble in water but soluble in organic solvents only. Solubility of smaller ketones and aldehydes are result of the hydrogen bond formation between the lone pairs of oxygen atom and the hydrogen atom in the water molecule.
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Chapter 15 Solutions
GENERAL,ORGANIC,+BIO.CHEM.-MINDTAP
- Q1: Answer the questions for the reaction below: ..!! Br OH a) Predict the product(s) of the reaction. b) Is the substrate optically active? Are the product(s) optically active as a mix? c) Draw the curved arrow mechanism for the reaction. d) What happens to the SN1 reaction rate in each of these instances: 1. Change the substrate to Br "CI 2. Change the substrate to 3. Change the solvent from 100% CH3CH2OH to 10% CH3CH2OH + 90% DMF 4. Increase the substrate concentration by 3-fold.arrow_forwardExperiment 27 hates & Mechanisms of Reations Method I visual Clock Reaction A. Concentration effects on reaction Rates Iodine Run [I] mol/L [S₂082] | Time mo/L (SCC) 0.04 54.7 Log 1/ Time Temp Log [ ] 13,20] (time) / [I] 199 20.06 23.0 30.04 0.04 0.04 80.0 22.8 45 40.02 0.04 79.0 21.6 50.08 0.03 51.0 22.4 60-080-02 95.0 23.4 7 0.08 0-01 1970 23.4 8 0.08 0.04 16.1 22.6arrow_forward(15 pts) Consider the molecule B2H6. Generate a molecular orbital diagram but this time using a different approach that draws on your knowledge and ability to put concepts together. First use VSEPR or some other method to make sure you know the ground state structure of the molecule. Next, generate an MO diagram for BH2. Sketch the highest occupied and lowest unoccupied MOs of the BH2 fragment. These are called frontier orbitals. Now use these frontier orbitals as your basis set for producing LGO's for B2H6. Since the BH2 frontier orbitals become the LGOS, you will have to think about what is in the middle of the molecule and treat its basis as well. Do you arrive at the same qualitative MO diagram as is discussed in the book? Sketch the new highest occupied and lowest unoccupied MOs for the molecule (B2H6).arrow_forward
- Q8: Propose an efficient synthesis of cyclopentene from cyclopentane.arrow_forwardQ7: Use compound A-D, design two different ways to synthesize E. Which way is preferred? Please explain. CH3I ONa NaOCH 3 A B C D E OCH3arrow_forwardPredict major product(s) for the following reactions. Note the mechanism(s) of the reactions (SN1, E1, SN2 or E2).arrow_forward
- (10 pts) The density of metallic copper is 8.92 g cm³. The structure of this metal is cubic close-packed. What is the atomic radius of copper in copper metal?arrow_forwardPredict major product(s) for the following reactions. Note the mechanism(s) of the reactions (SN1, E1, SN2 or E2).arrow_forwardPredict major product(s) for the following reactions. Note the mechanism(s) of the reactions (SN1, E1, SN2 or E2).arrow_forward
- Q3: Rank the following compounds in increasing reactivity of E1 and E2 eliminations, respectively. Br ca. go do A CI CI B C CI Darrow_forwardQ5: Predict major product(s) for the following reactions. Note the mechanism(s) of the reactions (SN1, E1, SN2 or E2). H₂O דיי "Br KN3 CH3CH2OH NaNH2 NH3 Page 3 of 6 Chem 0310 Organic Chemistry 1 HW Problem Sets CI Br excess NaOCH 3 CH3OH Br KOC(CH3)3 DuckDuckGarrow_forwardQ4: Circle the substrate that gives a single alkene product in a E2 elimination. CI CI Br Brarrow_forward
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