13.6
תדירות ספיגה ב-IR: היברידיזציה
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Q1: Why does C–H stretching frequency increase with sp hybridization?
sp-hybridized carbon has maximum s character (50%), holding electrons closer to the nucleus. This creates shorter, stronger C–H bonds that vibrate at higher frequencies than sp2 and sp3 carbons. The trend reflects bond strength: sp C–H stretches around 3300 cm−1, sp2 around 3100 cm−1, and sp3 below 3000 cm−1.
Q2: How does s character affect C–H bond strength in different hybridizations?
Higher s character means electron density concentrates closer to the nucleus, strengthening the C–H bond. sp orbitals (50% s character) create the strongest bonds, followed by sp2 (33% s character) and sp3 (25% s character). Stronger bonds require more energy to stretch, producing higher IR absorption frequencies.
Q3: What IR frequencies characterize alkanes, alkenes, and alkynes?
Alkanes show C–H stretches below 3000 cm−1 due to sp3 hybridization. Alkenes display bands around 3100 cm−1 from sp2 C–H bonds. Alkynes exhibit stretches near 3300 cm−1 from sp C–H bonds. These distinct frequencies allow identification of hydrocarbon types in IR spectra.
Q4: Why don't tetrasubstituted alkenes show a C–H stretching band?
Tetrasubstituted alkenes lack C–H bonds entirely because all four positions on the double bond are occupied by carbon or substituent groups. Without C–H bonds present, no characteristic stretching absorption appears at 3100 cm−1 in the IR spectrum. This absence is diagnostic for tetrasubstituted alkenes.
Q5: What causes internal alkynes to lack the 3300 cm−1 IR band?
Internal alkynes have no terminal C–H bonds because both ends of the triple bond connect to carbon atoms. Since the 3300 cm−1 band corresponds to sp C–H stretching, internal alkynes without these bonds do not exhibit this characteristic absorption in the IR spectrum.
Q6: How can IR spectroscopy distinguish between sp, sp2, and sp3 carbons?
Each hybridization produces distinct C–H stretching frequencies: sp carbons absorb at 3300 cm−1, sp2 at 3100 cm−1, and sp3 below 3000 cm−1. By analyzing the IR spectrum in the 2600–3400 cm−1 range, chemists identify which carbon types are present in an unknown compound.
Q7: What role does orbital geometry play in C–H bond length and IR frequency?
sp orbitals orient linearly with maximum s character, creating the shortest C–H bonds. sp2 orbitals form trigonal geometry with less s character, producing longer bonds. sp3 orbitals adopt tetrahedral geometry with minimum s character, yielding the longest bonds. Shorter bonds stretch at higher frequencies.