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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
Bundle: General, Organic, and Biological Chemistry, 7th + OWLv2 Quick Prep for General Chemistry, 4 terms (24 months) Printed Access Card
- Curved arrows are used to illustrate the flow of electrons using the provided starting and product structures draw the curved electron pushing arrows for the following reaction or mechanistic steps Ether(solvent)arrow_forwardThis deals with synthetic organic chemistry. Please fill in the blanks appropriately.arrow_forwardUse the References to access important values if needed for this question. What is the IUPAC name of each of the the following? 0 CH3CHCNH₂ CH3 CH3CHCNHCH2CH3 CH3arrow_forward
- You have now performed a liquid-liquid extraction protocol in Experiment 4. In doing so, you manipulated and exploited the acid-base chemistry of one or more of the compounds in your mixture to facilitate their separation into different phases. The key to understanding how liquid- liquid extractions work is by knowing which layer a compound is in, and in what protonation state. The following liquid-liquid extraction is different from the one you performed in Experiment 4, but it uses the same type of logic. Your task is to show how to separate apart Compound A and Compound B. . Complete the following flowchart of a liquid-liquid extraction. Handwritten work is encouraged. • Draw by hand (neatly) only the appropriate organic compound(s) in the boxes. . Specify the reagent(s)/chemicals (name is fine) and concentration as required in Boxes 4 and 5. • Box 7a requires the solvent (name is fine). • Box 7b requires one inorganic compound. • You can neatly complete this assignment by hand and…arrow_forwardb) Elucidate compound D w) mt at 170 nd shows c-1 stretch at 550cm;' The compound has the ff electronic transitions: 0%o* and no a* 1H NMR Spectrum (CDCl3, 400 MHz) 3.5 3.0 2.5 2.0 1.5 1.0 0.5 ppm 13C{H} NMR Spectrum (CDCl3, 100 MHz) Solvent 80 70 60 50 40 30 20 10 0 ppm ppm ¹H-13C me-HSQC Spectrum ppm (CDCl3, 400 MHz) 5 ¹H-¹H COSY Spectrum (CDCl3, 400 MHz) 0.5 10 3.5 3.0 2.5 2.0 1.5 1.0 10 15 20 20 25 30 30 -35 -1.0 1.5 -2.0 -2.5 3.0 -3.5 0.5 ppm 3.5 3.0 2.5 2.0 1.5 1.0 0.5 ppmarrow_forwardShow work with explanation. don't give Ai generated solutionarrow_forward
- Redraw the flowchartarrow_forwardredraw the flowchart with boxes and molecules written in themarrow_forwardPart I. a) Elucidate the structure of compound A using the following information. • mass spectrum: m+ = 102, m/2=57 312=29 • IR spectrum: 1002.5 % TRANSMITTANCE Ngg 50 40 30 20 90 80 70 60 MICRONS 5 8 9 10 12 13 14 15 16 19 1740 cm M 10 0 4000 3600 3200 2800 2400 2000 1800 1600 13 • CNMR 'H -NMR Peak 8 ppm (H) Integration multiplicity a 1.5 (3H) triplet b 1.3 1.5 (3H) triplet C 2.3 1 (2H) quartet d 4.1 1 (2H) quartet & ppm (c) 10 15 28 60 177 (C=0) b) Elucidate the structure of compound B using the following information 13C/DEPT NMR 150.9 MHz IIL 1400 WAVENUMBERS (CM-1) DEPT-90 DEPT-135 85 80 75 70 65 60 55 50 45 40 35 30 25 20 ppm 1200 1000 800 600 400arrow_forward
- • Part II. a) Elucidate The structure of compound c w/ molecular formula C10 11202 and the following data below: • IR spectra % TRANSMITTANCE 1002.5 90 80 70 60 50 40 30 20 10 0 4000 3600 3200 2800 2400 2000 1800 1600 • Information from 'HAMR MICRONS 8 9 10 11 14 15 16 19 25 1400 WAVENUMBERS (CM-1) 1200 1000 800 600 400 peak 8 ppm Integration multiplicity a 2.1 1.5 (3H) Singlet b 3.6 1 (2H) singlet с 3.8 1.5 (3H) Singlet d 6.8 1(2H) doublet 7.1 1(2H) doublet Information from 13C-nmR Normal carbon 29ppm Dept 135 Dept -90 + NO peak NO peak 50 ppm 55 ppm + NO peak 114 ppm t 126 ppm No peak NO peak 130 ppm t + 159 ppm No peak NO peak 207 ppm по реак NO peakarrow_forwardCould you redraw these and also explain how to solve them for me pleasarrow_forwardNonearrow_forward
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